Problem: The equation of a circle $C$ is $x^2+y^2+16x-18y+136 = 0$. What is its center $(h, k)$ and its radius $r$ ?
Answer: To find the equation in standard form, complete the square. $(x^2+16x) + (y^2-18y) = -136$ $(x^2+16x+64) + (y^2-18y+81) = -136 + 64 + 81$ $(x+8)^{2} + (y-9)^{2} = 9 = 3^2$ Thus, $(h, k) = (-8, 9)$ and $r = 3$.